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I am building a photo sharing service in php. I am using a lightbox in jquery which will popup when we click 'add' button to add photos. We can upload multiple photos. Also I am using ajax to upload photos so that the page is not reloaded. I want that after I upload photos the same will be automatically loaded in my gallery and the gallery should display new photos without need to refresh the page. The photos will have a particular id for a particular user in the database, so ulitmately the change in the table for the user should be reflected. Now the problem is that I don't have control over the closing button of the lightbox. Therefore I cannot modify it to call any other function so that it performs the query and displays my photos using ajax. I have heard that we can detect changes in the database automatically using JSON, but I have never used JSON and know almost nothing of it. Can anyone illustrate a simple example in php of how can the changes in a mysql table detected using JSON? Is there any other way to achieve it? Please help me.

Sanks R
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  • Have the upload handler on the server return whatever's needed to your client-side javascript to identify the new images. Id numbers, html snippets, etc... which you then insert dynamically into the gallery. – Marc B May 12 '11 at 04:15

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JSON an ideal data-interchange language.JSON is used for fast data interaction only.so you can not do this with out help of DOM request. http://www.json.org/

You can do this with ajax. You make ajax request for every second for latest update.

or

use long polling method like comet

Implement COMET with PHP

EDIT:- gowri:How to make ajax request per second?

use setInterval

setInterval(function(){

//make your ajax call here

},1000);

Community
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Gowri
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  • How to make ajax request per second? Does it affect the interface? I mean does the user get a sense that the gallery is refreshed now and then? – Sanks R May 12 '11 at 04:35
  • check update answer. do you want to show the gallery is refreshing to user – Gowri May 12 '11 at 04:45
  • than no problem.use JSON and get records in array format,with that array draw your page. live example is here http://www.beezid.com/.use firfug to see the response – Gowri May 12 '11 at 04:57